Question Number 177194 by peter frank last updated on 02/Oct/22
$$\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{sin}\:\mathrm{x}\right)^{\frac{\mathrm{14}}{\mathrm{9}}} \left(\mathrm{cos}\:\mathrm{x}\right)^{\frac{\mathrm{4}}{\mathrm{9}}} } \\ $$
Answered by som(math1967) last updated on 02/Oct/22
$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\left({sinx}\right)^{\frac{\mathrm{14}}{\mathrm{9}}} ×\left({cosx}\right)^{\frac{\mathrm{4}}{\mathrm{9}}} ×{sec}^{\mathrm{2}} {x}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\left({sinx}\right)^{\frac{\mathrm{14}}{\mathrm{9}}} ×\left({cosx}\right)^{\frac{\mathrm{4}}{\mathrm{9}}−\mathrm{2}} } \\ $$$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\left({tanx}\right)^{\frac{\mathrm{14}}{\mathrm{9}}} } \\ $$$$\:\int\left({tanx}\right)^{−\frac{\mathrm{14}}{\mathrm{9}}} {d}\left({tanx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{14}}{\mathrm{9}}}\left({tanx}\right)^{\mathrm{1}−\frac{\mathrm{14}}{\mathrm{9}}} \:+{C} \\ $$$$=−\frac{\mathrm{9}}{\mathrm{5}\left({tanx}\right)^{\frac{\mathrm{5}}{\mathrm{9}}} }\:+{C} \\ $$
Commented by peter frank last updated on 02/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$